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Suppose $R$ is an ordered field. Call a continuous map $f: R \rightarrow R$ a contraction if there exists $r < 1$ (in $R$) such that $|f(x)-f(y)| \leq r |x-y|$ for all $x,y \in R$ (where $|x| := \max(x,-x)$). Suppose that every contraction from $R$ to $R$ has a unique fixed point. Must $R$ be the field of real numbers?

For a related question, see http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem .

Jacek Jachymski's article "A discrete fixed point theorem of Eilenberg as a particular case of the contraction principle" ( http://emis.impa.br/EMIS/journals/HOA/FPTA/2004/131.pdf ) and the references it contains may be relevant. However, the non-Archimedean metric spaces that the article considers are bounded, which non-Archimedean ordered fields certainly are not. Also, my question is not about metric spaces, since my notion of distance lives in $R$ itself, not the real numbers.

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# Converse to Banach’s fixed point theorem for ordered fields?

Suppose $R$ is an ordered field. Call a continuous map $f: R \rightarrow R$ a contraction if there exists $r < 1$ (in $R$) such that $|f(x)-f(y)| \leq r |x-y|$ for all $x,y \in R$ (where $|x| := \max(x,-x)$). Suppose that every contraction from $R$ to $R$ has a unique fixed point. Must $R$ be the field of real numbers?

For a related question, see http://mathoverflow.net/questions/26119/converse-to-banachs-fixed-point-theorem .